# Open set around intersection of two convex sets

Consider two closed (compact if needed) convex sets $$E$$ and $$F$$. Define the $$\epsilon$$ neighborhood of set $$E$$ as $$E_\epsilon = \cup_{x \in E} B_\epsilon (x),$$ where $$B_\epsilon(x)$$ is the open $$\epsilon$$-ball around $$x$$.

Now, given a $$\delta>0$$, is there always an $$\epsilon>0$$, such that $$E_\epsilon \cap F \subseteq (E\cap F)_\delta.$$

One counter example for $$E$$ not closed: The intersection $$E\cap F$$ is the top right corner, and its $$\delta$$-neighborhood is just the $$\delta$$-ball. However, for any $$\epsilon>0$$, $$E_\epsilon \cap F = F$$, which would easily be out of $$(E\cap F)_\delta$$.

Some backgrounds: I started with the problem without the closedness and convexity assumption, and intuitively it makes sense. Then I started to see counter examples. With (almost) the strongest assumption, is the above statement correct?

One more assumption I can think of is to assume that $$\mathrm{int}(E) \cap \mathrm{int}(F) \ne \emptyset$$.

Thanks for any help / suggestions / hints!

• You are assuming that $E\cap F\neq \emptyset$, right? Otherwise, its $\delta$-neighbourhood would not be defined. Aug 15, 2022 at 18:23
• Also, is this question in $\mathbb{R}^d$? Or in a more general vector space? Aug 15, 2022 at 18:26
• Hey Keen, yes, I am assuming $E \cap F$ is non-empty. And yes, you can assume $\mathbb{R}^d$. Thanks for checking my question. Aug 15, 2022 at 20:40

Assuming $$X=\mathbb{R}^d$$, I think the answer is true. Assume towards contradiction that the statement is not true. Then for every $$n$$ there exists an $$x_n\in F$$ such that $$d(x_n,E)<\frac{1}{n} \quad \text{and} \quad d(x_n,E\cap F)\geq \delta.$$

If $$F$$ is compact there exists an $$x_\infty \in F$$ such that $$x_n\to x_\infty$$. Since $$E$$ is closed and $$d(x_n,E)\to 0$$, we have $$x_\infty \in E$$. Hence, $$x_\infty \in E\cap F$$ while $$d(x_\infty,E\cap F)\geq \delta$$. This is clearly a contradiction.

My proof is weird because it just needs for the intersection not to be empty and the sets to be closed. Nowhere do I rely on convexity.

• Hey Keen, thanks a lot for answering the question. The only spot I would make change to your post is $x_{n_k} \to x_\infty$, we probably only have subsequence convergence for compact set. Other than this, everything looks right. Thank you so much. Aug 15, 2022 at 21:28

If we allow $$E$$ and $$F$$ to be compact, and $$E \cap F \neq \emptyset$$ then the result is quite trivial in any metric space. In fact we don't need compactness - if one of them is bounded statement holds.

If $$E$$ was bounded, so would be $$E_\epsilon$$, so $$E_\epsilon \cap F$$ is bounded either way (as a subset of bounded set). Choose ball $$B(x, r) \supseteq E_\epsilon \cap F$$ and any point $$p \in E\cap F$$. For $$\delta = r + d(x, p)$$ we have $$B(x, r) \subseteq B(p, \delta) \subseteq (F\cap E)_\delta$$.

So sets don't even need to be closed.

• Hey Esgeriath, thanks a lot for trying to answer my question. I think maybe I did not pose the question well enough. I was asking given an arbitrary $\delta$ if there exists an $\epsilon$ that will satisfy $E_\epsilon \cap F \subseteq (E\cap F)_\delta$. I added a counter-example in the original post, and I hope it would clarify a little bit. Again thanks for the help! Aug 15, 2022 at 21:19
• @Lyapunov1729 so you're asking about what conditions would need to hold in order to ensure your statement? I mean, you just gave counterexample yourself Aug 15, 2022 at 21:45
• In the counter example, the set $E$ is not closed. That's why I added closed / compact assumption to the problem statement. Aug 15, 2022 at 21:52